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  1. #1
    Forum Admin topsquark's Avatar
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    Fields

    Got another one for y'all. I got part of the way through this, but I'm stuck for a way to continue.

    Here's the question:

    "Let K be a field, and let R be a subring of K. Suppose that every element of K satisfies a polynomial whose coefficients are in R and whose highest degree coefficient is 1. Show that R is actually a subfield of K."

    The question goes on to give a couple of hints:
    1) Show that R is an integral domain
    2) Given a nonzero $\displaystyle a \epsilon R$ we know that 1/a satisfies a suitable polynomial with coefficients in R. Manipulate this polynomial.

    Here's what I've got so far:

    Proof: R is an integral domain.
    Assume R is a subring of the field K. Then R is an integral domain because any element of R is also an element of K, thus the subring R is commutative and contains no zero-divisors. (End of proof.)

    For the second part, here's as far as I've gotten.

    By hypothesis, $\displaystyle \forall a\epsilon K$ $\displaystyle \exists \{r_i\} \subset R$ and $\displaystyle n \epsilon Z_{+} $ such that:
    $\displaystyle a^n+r_{n-1}a^{n-1}+...+r_0=0$

    Now, since $\displaystyle a \epsilon K$ and K is a field $\displaystyle \exists 1/a \epsilon K$. We may multiply the above equation by $\displaystyle (1/a)^n$:
    $\displaystyle 1+r_{n-1}(1/a)+...+r_0(1/a)^n=0$
    or
    $\displaystyle r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$.

    Again, since $\displaystyle a \epsilon K$ and K is a field $\displaystyle \exists 1/a \epsilon K$, by hypothesis:
    $\displaystyle (1/a)^m+q_{m-1}(1/a)^{m-1}+...+q_0=0$; $\displaystyle \{q_i\} \subset R$
    where m does not need to be equal to n. We may multiply this equation by $\displaystyle r_0$:
    $\displaystyle r_0(1/a)^m+r_0q_{m-1}(1/a)^{m-1}+...+r_0q_0=0$.
    Compare this with
    $\displaystyle r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$.

    I stated before that m was not necessarily equal to n, but these equations together show that the set of all possible values for m (given this a) contains the value n. So set m=n. Furthermore, R contains no zero-divisors, so the second equation is not trivial (that is, all of the coefficients are non-zero.) Finally, the leading coefficient is the same in both equations. Thus I conclude the polynomials are equal in the sense that ALL the coefficients of like terms are equal.

    In particular, comparison of the constant term shows $\displaystyle r_0q_0=1$ where both $\displaystyle r_0,q_0 \epsilon R$. Thus, by definition, $\displaystyle q_0$ is the multiplicative inverse of $\displaystyle r_0$.

    This is as far as I've gotten. I can show by this method that some elements of R have inverses in R, but not all of them. Only those $\displaystyle r_0$ values that come up in polynomials for given values of a will have inverses, not the whole set R.

    Any thoughts to finish this, or am I beating down a dead end?

    Thanks!
    -Dan
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark
    Got another one for y'all. I got part of the way through this, but I'm stuck for a way to continue.

    Here's the question:

    "Let K be a field, and let R be a subring of K. Suppose that every element of K satisfies a polynomial whose coefficients are in R and whose highest degree coefficient is 1. Show that R is actually a subfield of K."

    The question goes on to give a couple of hints:
    1) Show that R is an integral domain
    2) Given a nonzero $\displaystyle a \epsilon R$ we know that 1/a satisfies a suitable polynomial with coefficients in R. Manipulate this polynomial.

    Here's what I've got so far:

    Proof: R is an integral domain.
    Assume R is a subring of the field K. Then R is an integral domain because any element of R is also an element of K, thus the subring R is commutative and contains no zero-divisors. (End of proof.)

    For the second part, here's as far as I've gotten.

    By hypothesis, $\displaystyle \forall a\epsilon K$ $\displaystyle \exists \{r_i\} \subset R$ and $\displaystyle n \epsilon Z_{+} $ such that:
    $\displaystyle a^n+r_{n-1}a^{n-1}+...+r_0=0$

    Now, since $\displaystyle a \epsilon K$ and K is a field $\displaystyle \exists 1/a \epsilon K$. We may multiply the above equation by $\displaystyle (1/a)^n$:
    $\displaystyle 1+r_{n-1}(1/a)+...+r_0(1/a)^n=0$
    or
    $\displaystyle r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$.

    Again, since $\displaystyle a \epsilon K$ and K is a field $\displaystyle \exists 1/a \epsilon K$, by hypothesis:
    $\displaystyle (1/a)^m+q_{m-1}(1/a)^{m-1}+...+q_0=0$; $\displaystyle \{q_i\} \subset R$
    where m does not need to be equal to n. We may multiply this equation by $\displaystyle r_0$:
    $\displaystyle r_0(1/a)^m+r_0q_{m-1}(1/a)^{m-1}+...+r_0q_0=0$.
    Compare this with
    $\displaystyle r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$.

    I stated before that m was not necessarily equal to n, but these equations together show that the set of all possible values for m (given this a) contains the value n. So set m=n. Furthermore, R contains no zero-divisors, so the second equation is not trivial (that is, all of the coefficients are non-zero.) Finally, the leading coefficient is the same in both equations. Thus I conclude the polynomials are equal in the sense that ALL the coefficients of like terms are equal.

    In particular, comparison of the constant term shows $\displaystyle r_0q_0=1$ where both $\displaystyle r_0,q_0 \epsilon R$. Thus, by definition, $\displaystyle q_0$ is the multiplicative inverse of $\displaystyle r_0$.

    This is as far as I've gotten. I can show by this method that some elements of R have inverses in R, but not all of them. Only those $\displaystyle r_0$ values that come up in polynomials for given values of a will have inverses, not the whole set R.

    Any thoughts to finish this, or am I beating down a dead end?

    Thanks!
    -Dan
    I was finally able to figure it out. I'll post it for the sake of completeness.

    Our story so far...

    The subring R of the field K is actually an integral domain, so R contains no zero-divisors. Additionally, the hypothesis assures us that $\displaystyle \forall a \in K$ we have that $\displaystyle \exists \{r_i \} \subset R$ and $\displaystyle n \in \mathbb{Z}_{+}$ such that:
    $\displaystyle a^n+r_{n-1}a^{n-1}+...+r_0=0$.
    This leads to the result that $\displaystyle r_0$ has a multiplicative inverse in R.

    The problem at this point is that we need to prove that an inverse exists for all elements of R. I was finally able to get around that problem:

    Since $\displaystyle a^n+r_{n-1}a^{n-1}+...+r_0=0$, take any $\displaystyle b \in R$ and form the product polynomial:
    $\displaystyle (a+b)(a^n+r_{n-1}a^{n-1}+...+r_0)=0$
    This produces a new polynomial:
    $\displaystyle a^{n+1}+...+br_0=0$.
    By the construction in the previous post (and as repeated above) the existence of this polynomial implies that the element $\displaystyle br_0$ has an inverse in R also. Recall that multiplication in R is commutative (since it is a subring of the field K.) Thus:
    $\displaystyle \frac{1}{br_0} \in R$ $\displaystyle \Rightarrow$ $\displaystyle r_0 \frac{1}{br_0} \in R$ $\displaystyle \Rightarrow$ $\displaystyle \frac{1}{b} \in R$ $\displaystyle \forall b \in R$, since multiplication in R is closed by definition.

    So all elements of R have a multiplicative inverse, making the integral domain R a field.

    -Dan
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  3. #3
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    I think you can make it rather shorter. For $\displaystyle r \in R$, the inverse $\displaystyle r^{-1} \in K$ satisfies a monic equation $\displaystyle r^{-n} + a_{n-1}r^{-n+1} + \cdots + a_1 r^{-1} + a_0 = 0$ with all the $\displaystyle a_i \in R$. Multiply by $\displaystyle r^{n-1}$ to get $\displaystyle r^{-1} = -a_{n-1} - \cdots - a_0 r^{n-1}$ and the RHS is in $\displaystyle R$ since it is a ring.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rgep
    I think you can make it rather shorter. For $\displaystyle r \in R$, the inverse $\displaystyle r^{-1} \in K$ satisfies a monic equation $\displaystyle r^{-n} + a_{n-1}r^{-n+1} + \cdots + a_1 r^{-1} + a_0 = 0$ with all the $\displaystyle a_i \in R$. Multiply by $\displaystyle r^{n-1}$ to get $\displaystyle r^{-1} = -a_{n-1} - \cdots - a_0 r^{n-1}$ and the RHS is in $\displaystyle R$ since it is a ring.
    Bugger.

    I hate it when I miss the easy way...

    Thanks for the tip!

    -Dan
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